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# hw27b - aljabr(faa335 Hw27 Ross(89251 This print-out should...

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aljabr (faa335) – Hw27 – Ross – (89251) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Pure helium gas is admitted into a tank con- taining a movable piston. The initial vol- ume, pressure, and temperature of the gas are 0 . 047 m 3 , 197 kPa, and 421 K. If the volume is decreased to 0 . 011 m 3 and the pressure is increased to 480 kPa, find the final temperature of the gas. Correct answer: 240 . 078 K. Explanation: Given : V i = 0 . 047 m 3 , P i = 197 kPa , T i = 421 K , V f = 0 . 011 m 3 , and P f = 480 kPa . If no gas escapes from the tank, the number of moles remains constant; therefore, using the equation P V = n R T at the initial and final points gives P i V i T i = P f V f T f T f = P f V f T i P i V i = (480 kPa) ( 0 . 011 m 3 ) (421 K) (197 kPa ) (0 . 047 m 3 ) = 240 . 078 K . 002 (part 1 of 2) 10.0 points A sphere 30 cm in diameter contains an ideal gas at 1 . 2 atm and 15 C. As the sphere is heated to 119 C, gas is allowed to escape. The valve is closed and the sphere is placed in an ice-water bath. How many moles of gas escape from the sphere as it warms? Correct answer: 0 . 190402 mol. Explanation: Given : r = 30 cm 2 = 0 . 15 m , P = 1 . 2 atm , T 1 = 15 C = 288 K , T 2 = 119 C = 392 K , and R = 8 . 31451 J / K · mol . The volume of the sphere is V = 4 π r 3 3 = 4 π (0 . 15 m) 3 3 = 0 . 0141372 m 3 .

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hw27b - aljabr(faa335 Hw27 Ross(89251 This print-out should...

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